When entered into Wolfram Alpha, $\infty^\infty$ results in complex infinity.

Why is $\infty^\infty=\tilde\infty$?

  • 31
    $\begingroup$ It isn't. $\infty^\infty$ is meaningless. Wolfram Alpha isn't truth, and you'd have to ask one of the programmers why their software chooses to respond that. $\endgroup$ – Thomas Andrews Dec 31 '16 at 19:08
  • 8
    $\begingroup$ Just an advice: dont trust wolfram-alpha or any other mathematical software. $\endgroup$ – Masacroso Dec 31 '16 at 19:10
  • 2
    $\begingroup$ There may be some computational purposes behind. For example, Series[(1+1/n)^n,{n,infinity,2}] gives $$e-\frac{e}{2n}+O\left( \frac{1}{n^2} \right)$$ $\endgroup$ – Ng Chung Tak Dec 31 '16 at 19:26
  • 2
    $\begingroup$ @Canardini That makes sense on the Riemann sphere, though. $\endgroup$ – Akiva Weinberger Dec 31 '16 at 19:34
  • 5
    $\begingroup$ I'm glad those with understanding answered this question, when so many were quick to say it was meaningless and an error in software. $\endgroup$ – user304051 Jan 1 '17 at 4:07

WA's ComplexInfinity is the same as Mathematica's: it represents a complex "number" which has infinite magnitude but unknown or nonexistent phase. One can use DirectedInfinity to specify the phase of an infinite quantity, if it approaches infinity in a certain direction. The standard Infinity is the special case of phase 0. Note that Infinity is different from Indeterminate (which would be the output of e.g. 0/0).

Some elucidating examples:

  • 0/0 returns Indeterminate, since (for instance) the limit may be approached as $\frac{1/n}{1/n}$ or $\frac{2/n}{1/n}$, resulting in two different real numbers.
  • 1/0 returns ComplexInfinity, since (for instance) the limit may be approached as $\frac{1}{-1/n}$ or as $\frac{1}{1/n}$, but every possible way of approaching the limit gives an infinite answer.
  • Abs[1/0] returns Infinity, since the limit is guaranteed to be infinite and approached along the real line in the positive direction.

In your particular example, you get ComplexInfinity because the infinite limit may be approached as (e.g.) $n^n$ or as $n^{n+i}$.

  • 13
    $\begingroup$ Glad to see this as an answer. Well addresses Mathematica's reasoning here, with well drawn out examples. +1. $\endgroup$ – Brevan Ellefsen Dec 31 '16 at 19:41
  • $\begingroup$ I doubt $(-n)^n$ is a valid way to do the limit — presumably WA interprets the "infinity" in the question to be a directed infinity, of argument/phase zero (that is, "positive infinity"). However, $n^{n+i}$ would work. $\endgroup$ – Akiva Weinberger Dec 31 '16 at 19:46
  • 6
    $\begingroup$ 0/0, outside of a limit, is not indeterminate: it is simply undefined. (Although we might say the lack of a definition is motivated by no consistent limits in this regard.) Indeed, Wolfram Alpha returns "undefined", not "indeterminate", in this case. But then it appends some indeterminate limits in case you want to investigate those, which likely furthers the confusion. $\endgroup$ – Daniel R. Collins Jan 1 '17 at 0:26
  • 1
    $\begingroup$ Thanks, this answer along with Brevan Ellefsen's answer explain it very clearly! If I could, I would accept both as the answer. I have learned a lot through both the answers and comments on this question, and this discussion is an exemplary example of the awesome culture of the Math Stack Exchange! $\endgroup$ – esote Jan 1 '17 at 1:14
  • 1
    $\begingroup$ @Idempotence thank you for such a kind comment. A compliment like that is always better than the reputation from accepting or upvoting in my book! I hope you find the rest of your time on this site equally enjoyable, and if you really enjoy the site, I invite you to try answering questions sometime! I love answering questions on this site for much the same reason you enjoyed the answers you got - the supportive community and the learning I do along the way. I imagine I often end up learning a whole lot more from the posts I answer than the OP does! $\endgroup$ – Brevan Ellefsen Jan 1 '17 at 17:18

TLDR: $\infty$ is not a number, and thus $\infty^\infty$ is meaningless, and Wolfram Alpha is using $\tilde\infty$ to represent something I like to to think of in the sense of "one-point compactification", a topological concept

A.) The symbol $\infty$ is not a number in its own right. It can represent a lot of things, and many different objects can be "infinitely large". Just think of something infinite as something not finite and you are generally off to a good start.

B.) If infinity is not a number, we can't do arithmetic on it that makes sense in all context, and so we most definitely can't exponetiate it meaningfully without some fundamentals first. For example,

  1. there are some infinite things where arithmetic is defined (in some sense - check out ordinal numbers),
  2. There are some things where basic arithmetic does nothing (i.e. $\infty+1=\infty$)
  3. There are some infinite thing where you most definitely cannot do arithmetic in the way you are used to (take, for example, the limit concept of infinity from a basic Calculus class)

C) Wolfram Alpha appears to represent a lot of things as $\tilde\infty$ that are ill-defined according to the real-number system you are used to - for example, according to Wolfram Alpha, $\frac{1}{0}=\tilde\infty$, whereas I would say that $\frac{1}{0}$ is undefined. You could stretch this to say that $\frac{1}{0} = \lim_{x\to 0} \frac 1x = \pm\infty$ in the extended-real number system, but this is starting to push things. To really understand what Wolfram Alpha is doing you must first understand the one-point compactification of $\Bbb C$. See my note at the bottom for links and more details.


  1. It has been pointed out in the comments that I ought to mention that Mathematica/WA use $\tilde\infty$ to represent an infinite magnitude number with no defined phase. This seemed more software dependent, though I get where the the commentator is coming from. I chose to focus on the real number system the OP is likely accustomed to, and focus on the concept of $\infty^\infty$ itself, not the software's interpretation.

  2. $\tilde\infty$ does have meaning in some sense - see for example, this Wikipedia page on "one-point compactification". Just imagine the complex plane as a big sheet where we take the edges and pull them all into one point and call it $\infty$. For more information look at the page on the Riemann Sphere. Similarly, $\infty$ can have a defined meaning in certain contexts outside of the scope of this question - see, for example, the Wikipedia page on the Real Projective Line. In these contexts, one could perhaps consider $\infty$ a number. For more on this, see the Wikipedia page "Projectively Extended Real Line".

  • 2
    $\begingroup$ Quibble: It's not a real number. What about the point of $RP^1$ with coordinates $(0,1)$ ? $\endgroup$ – Spencer Dec 31 '16 at 19:26
  • 1
    $\begingroup$ The points of $\mathbb RP^1$ are not numbers. @Spencer $\endgroup$ – Thomas Andrews Dec 31 '16 at 19:27
  • 1
    $\begingroup$ @PatrickStevens I agree with you completely.... I edited my post to reflect this with a note at the end. I attempted to more explain the concept of $\infty^\infty$ and what it could mean instead of what $\tilde\infty$ means according to Mathematica... feel free to post an answer tackling that part of the question! $\endgroup$ – Brevan Ellefsen Dec 31 '16 at 19:33
  • 1
    $\begingroup$ @ThomasAndrews Oh really? One of the answers at math.stackexchange.com/questions/494854/what-is-a-number contradicts this. We could argue at length about what a number is, but it would be beyond the scope of this question. $\endgroup$ – Spencer Dec 31 '16 at 19:35
  • 1
    $\begingroup$ I am sure $\tilde\infty$ refers to the extra point in the one-point compactification of $\Bbb C$ (that is, the Riemann sphere). $\endgroup$ – Akiva Weinberger Dec 31 '16 at 19:43

An interpretation in $\overline{\mathbb{C}}=\mathbb{C}\cup \{\infty\}$ of $\infty^\infty$ is via limits. For example $$\lim_{z\to 0}\left(\frac{1}{|z|}\right)^{\frac{1}{|z|}}\underbrace{=}_{\text{symbolic equality}}\infty^\infty=\infty.$$ Note. $-\infty$ and $+\infty$ don't belong to $\overline{\mathbb{C}}$.


As far as I can tell: When $x$ and $y$ approach positive infinity, Wolfram Alpha assumes they may do so through the complex plane, as long as the arguments (angles) of $x$ and $y$ approach zero.

The argument of $x^{x+i}$ does not converge to any value as $x\to\infty$, even though the arguments of $x$ and $x+i$ go to zero. (This is because the argument is $\ln x$.) Thus, Wolfram Alpha responds with complex infinity (unknown argument) rather than positive infinity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.